Prove that if $f$ is integrable on $[0,1]$, then $\lim_{n→∞}\int_{0}^{1} x^{n}f(x)dx = 0$. 
Prove that if $f$ is integrable on $[0,1]$, then $\lim_{n→∞}\int_{0}^{1} x^{n}f(x)dx = 0$.

Attempt: this exercise has two parts. I did part a) already. From part a) we know that $g_{n} \geq 0$ is a sequence on integrable functions which satisfies  $\lim_{n→∞}\int_{a}^{b} g_{n}dx = 0$.
Then suppose $f$ is integrable, thus, it is bounded so let $m$ be the infimum and $M$ be the supremum of $f$ on the interval $[a,b]$.
Then by the Mean Value Theorem there is a $c \in [m, M]$ such that  $\int_{a}^{b} f(x)g(x)dx = c \int_{a}^{b} g(x)dx$.
Then let $g_{n} = x^{n} \geq 0$ $\forall x \in [0,1]$ Then $\int_{a}^{b} f(x)g_{n}(x)dx = c \int_{a}^{b} g_{n}(x)dx = c \int_{a}^{b} x^{n}dx$.
Can someone please help me continue?Any feedback/help would be appreciated. Thank you.
 A: If you know that $f$ is bounded, so that $-M \le f \le M$, then you have $\int_0^1 x^n f(x)\,dx \le M \int_0^1 x^n\,dx$.  But you can compute $\int_0^1 x^n\,dx$ directly and show that it converges to 0.  Likewise, $\int_0^1 x^n f(x)\,dx \ge -M \int_0^1 x^n\,dx$.  Use the squeeze theorem.
Alternatively, use the fact that $$\left| \int_0^1 x^n f(x)\,dx \right| \le \int_0^1 x^n |f(x)|\,dx \le M \int_0^1 x^n\,dx.$$
(This assumes that "integrable" means "Riemann integrable".  If it means "Lebesgue integrable", then integrable functions need not be bounded, and this proof doesn't work.  But if you're working with Lebesgue integrable functions then you probably know the dominated convergence theorem and should just use that.)
(Acknowledgement of priority: I just noticed that Jon's comment contains exactly the same hint.)
A: Edit: This is probably a better answer to the hypothetical question: "What is a problem that is trivially solved by the dominated convergence theorem, but requires substantially more work to do by hand."
One argument might be as follows, though maybe this uses more measure theory than what you want. The idea is that $f$ is almost bounded.
For any $\epsilon > 0$, there is an $M$ so that $f \leq M$ on a set $E$ with $\int_{E^c} f < \epsilon$. This is just because $[0,1] = \bigcup \{f \leq N\}$, plus absolute continuity of the measure defined by $f$. On $E$ the statement follows because $x^n \to 0$ in norm, so $\int_E x^n f \leq M \int_E x^n \to 0$ - say that for $n \geq N$, $\int_E x^n f \leq \epsilon$. Off of $E$, since $x^n \leq 1$ on $[0,1]$, we have that $\int_{E^c} x^n f \leq \int_{E^c} f \leq \epsilon$.
Putting this together, we have that $\int_{[0,1]} x^n f \leq \int_E x^n f + \int_{E^c} x^n f \leq 2 \epsilon$, when $n \geq N$.
